2022 AMC 10A Problems/Problem 4: Difference between revisions
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<math>\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}</math> | <math>\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}</math> | ||
== Solution == | == Solution 1 == | ||
The formula for fuel efficiency is <cmath>\frac{\text{Distance}}{\text{Gas Consumption}}.</cmath> | The formula for fuel efficiency is <cmath>\frac{\text{Distance}}{\text{Gas Consumption}}.</cmath> | ||
Note that <math>1</math> mile equals <math>\frac 1m</math> kilometers. We have <cmath>\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.</cmath> | Note that <math>1</math> mile equals <math>\frac 1m</math> kilometers. We have <cmath>\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.</cmath> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
== Solution 2 == | |||
Since it can be a bit odd to think of "liters per 100 km", this statement's numerical value is equivalent to <math>100\cdot\text{km per 1 liter}</math>. | |||
<math>1\text{km}=l\text{ liters},</math> so the numerator is simply <math>l</math>. Since <math>l</math> liters <math>=</math> <math>1</math> gallon, and <math>x</math> miles = <math>1</math> gallon, <math>1 liter = \frac{x}{l}</math>. | |||
Therefore, the requested expression is | |||
<cmath>100\cdot\frac{m}{(\frac{x}{l})} = \frac{100lm}{x} \rightarrow \boxed{\textbf{(E) } \frac{100lm}{x}}.</cmath> | |||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== | ||
Revision as of 22:54, 13 November 2022
Problem
In some countries, automobile fuel efficiency is measured in liters per 
kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals 
miles, and 
gallon equals 
liters. Which of the following gives the fuel efficiency in liters per kilometers for a car that gets 
miles per gallon?
Solution 1
The formula for fuel efficiency is ![\[\frac{\text{Distance}}{\text{Gas Consumption}}.\]](http://latex.artofproblemsolving.com/9/0/e/90eeec553fea9ce3012153b8582c2d9f6a593332.png)
Note that mile equals 
kilometers. We have ![\[\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.\]](http://latex.artofproblemsolving.com/6/0/c/60c6bc79b632eb92edc1ac2b7389e21da8beb47b.png)
Therefore, the answer is 
~MRENTHUSIASM
Solution 2
Since it can be a bit odd to think of "liters per 100 km", this statement's numerical value is equivalent to 
.
so the numerator is simply . Since liters 
gallon, and miles = gallon, 
.
Therefore, the requested expression is
![\[100\cdot\frac{m}{(\frac{x}{l})} = \frac{100lm}{x} \rightarrow \boxed{\textbf{(E) } \frac{100lm}{x}}.\]](http://latex.artofproblemsolving.com/e/4/e/e4ee1069494e72df91c7ecccb8e3578f17da353b.png)
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
